Problem: In the following diagram, $AB=50$.  Find $AX$.

[asy]
import markers;

real t=.67;
pair A=(0,0);
pair B=(3,-2);
pair C=(1.5,1.5);
pair X=t*A+(1-t)*B;

draw(C--A--B--C--X);

label("$A$",A,SW);
label("$B$",B,E);
label("$C$",C,N);
label("$X$",X,SW);

markangle(n=1,radius=15,A,C,X,marker(markinterval(stickframe(n=1),true)));
markangle(n=1,radius=15,X,C,B,marker(markinterval(stickframe(n=1),true)));

//label("$24$",.5*(B+X),SE);
label("$56$",.5*(B+C),E);
label("$28$",.5*(A+C),NW);

[/asy]
Explanation: The Angle Bisector Theorem tells us that  \[\frac{AX}{AC}=\frac{BX}{BC}\] so cross multiplying and substituting tells us  \[56AX=28BX\] or $BX=2AX$.

We want to find $AX$, so we write \[50=AB=AX+XB=AX+2AX=3AX.\] Solving gives us $AX=\boxed{\frac{50}3}$.